I have a problem where, given an array of integers, I need to find sets of three numbers that add up to equal zero. The below solution works but isn't as optimal as I'd like and I am looking for ways to optimize it to avoid unnecessary processing.
What I am doing below is I am iterating through the all combinations of numbers while eliminating iterating through the same indices in each nested loop and I am checking if the three numbers in the inner most loop add up to zero. If yes, I am converting the array to a string and if the string isn't already in the results array I am adding it. Right before returning I am then converting the strings back to an array.
I appreciate any suggestions on how to further optimize this or if I missed out on some opportunity to implement better. I am not looking for a total refactor, just some adjustments that will improve performance.
var threeSum = function(nums) {
const sorted = nums.sort();
if(sorted.length && (sorted[0] > 0 || sorted[sorted.length-1] < 0)) {
return [];
}
let result = [];
for(let i=0; i < sorted.length; i++) {
for(let z=i+1; z < sorted.length; z++) {
for(let q=z+1; q < sorted.length; q++) {
if(sorted[i]+sorted[z]+sorted[q] === 0) {
const temp = [sorted[i], sorted[z], sorted[q]].join(',');
if(!result.includes(temp)) {
result.push(temp);
}
}
}
}
}
return result.map(str => str.split(','));
};
Sample Input: [-1,0,1,2,-1,-4]
Expected Output: [[-1,-1,2],[-1,0,1]]
One obvious optimisation is to precalculate the sum of the two first numbers just before the third nested loop. Then compare in the third loop if that number equals the opposite of the third iterated number.
Second optimisation is to take advantage of the fact that your items are sorted and use a binary search for the actual negative of the sum of the two first terms in the rest of the array instead of the third loop. This second optimisation brings complexity from O(N3) down to O(N2LogN)
Which leads to the third optimisation, for which you can store in a map the sum as key and as value, an array of the different pairs which sum to the sum so that each time you want to operate the binary search again, first you check if the sum already exists in that map and if it does you can simply output the combination of each pair found at that sum’s index in the map coupled with the negative sum.
The OP's solution runs in O(N³) time with no additional storage.
The classic "use a hash table" solution to find the missing element can bring that down to O(N²) time with O(N) storage.
The solution involves building a number map using an object. (You could use a Map object as well, but then you can't be as expressive with ++ and -- operators). Then just an ordinary loop and inner loop to evaluate all the pairs. For each pair, find if the negative sum of those pairs is in the map.
function threeSum(nums) {
var nummap = {}; // map a value to the number of ocurrances in nums
var solutions = new Set(); // map of solutions as strings
// map each value in nums into the number map
nums.forEach((val) => {
var k = nummap[val] ? nummap[val] : 0; // k is the number of times val appears in nummap
nummap[val] = k+1; // increment by 1 and update
});
// for each pair of numbers, see if we can find a solution the number map
for (let i = 0; i < nums.length; i++) {
var ival = nums[i];
nummap[ival]--;
for (let j = i+1; j < nums.length; j++) {
var jval = nums[j];
nummap[jval]--;
var target = -(ival + jval); // this could compute "-0", but it works itself out since 0==-0 and toString will strip the negative off
// if target is in the number map, we have a solution
if (nummap[target]) {
// sort this three sum solution and insert into map of available solutions
// we do this to filter out duplicate solutions
var tmp = [];
tmp[0] = ival;
tmp[1] = jval;
tmp[2] = target;
tmp.sort();
solutions.add(tmp.toString());
}
nummap[jval]++; // restore original instance count in nummap
}
nummap[ival]--;
}
for (s of solutions.keys()) {
console.log(s);
}
}
threeSum([9,8,7,-15, -9,0]);
var threeSum = function(unsortedNums) {
const nums = unsortedNums.sort();
if(nums.length && (nums[0] > 0 || nums[nums.length-1] < 0)) {
return [];
}
const result = new Map();
for(let i=0; i < nums.length; i++) {
for(let z=i+1; z < nums.length; z++) {
for(let q=z+1; q < nums.length; q++) {
if(nums[i]+nums[z]+nums[q] === 0) {
const toAdd = [nums[i], nums[z], nums[q]];
const toAddStr = toAdd.join(',');
if(!result.has(toAddStr)) {
result.set(toAddStr, toAdd);
}
}
}
}
}
return Array.from(result.values());
};
Related
Example :
Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums
being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
My code to solve this problem is
var removeDuplicates = function(nums) {
const non_duplicates = [];
for (let i=0;i<=nums.length-1;i++){
if(!(non_duplicates.includes(nums[i]))){
non_duplicates.push(nums[i]);
}
}
console.log(non_duplicates)
return non_duplicates.length;
};
That console.log(non_duplicates) displays correct output in stdOut. But when I return non_duplictes it prints empty array in output. And when I return non_duplictes.length it returns some array rather than the length of that array.
Please don't suggest any other method to solve this problem. Just tell what is wrong with this code
You can see that problem here
Your solution doesn't modify the array inplace. That's the first problem.
A bigger problem is that your algorithm is with O(N^2) time complexity, due to includes call and given that "1 <= nums.length <= 3 * 104" your solution would be incredibly slow if it ever passes the tests.
The goal is to find a linear solution of the problem. One possible solution would be:
var removeDuplicates = function(nums) {
let j = 0;
for (let i = 1, len = nums.length; i < len; i++) {
if (nums[i] !== nums[j]) {
nums[++j] = nums[i];
}
}
return j + 1;
};
You can achieve that by using Set() Object.
let nums = [0,0,1,1,1,2,2,3,3,4];
const uniqueNums = [...new Set(nums)];
console.log(uniqueNums.length, uniqueNums);
You're returning the correct value but you are not overwriting the nums array with the non-duplicate values.
After calling your function, the return value should be 5 and the first 5 values in the passed-in nums array should be 0,1,2,3,4 (the value of the others doesn't matter).
Simplest solution that's compatible with your attempted solution is:
copy the values from non_duplicates into nums
return non_duplicates.length.
Here's a simple example of how to do this:
var removeDuplicates = function (nums) {
const non_duplicates = [];
for (let i = 0; i < nums.length; i++) {
if (!non_duplicates.includes(nums[i])) {
non_duplicates.push(nums[i]);
}
}
for (let i = 0; i < non_duplicates.length; i++) {
nums[i] = non_duplicates[i];
}
return non_duplicates.length;
};
Note: this is not the most efficient solution, and efficiency becomes important with the medium to hard Leetcode challenges, but it's simple, close to your intended solution, and it works.
I've got working code, but I'm looking to improve the way I understand and implement different coding techniques as a whole. I thought this problem presented a good chance to get feedback on what I'm doing.
The idea here is to take two arguments, an array and an integer. Identify all pairs in the array that sum to make the integer argument, and then return the sum of the indices. You cannot reuse an index, and you must always use the smallest index available to you. There is an explanation on the FCC code guide here: https://www.freecodecamp.org/learn/coding-interview-prep/algorithms/pairwise
So - here is the question. Is there any good way of doing this without using nested for loops? I am becoming increasingly aware of time/space complexities, and I know that O(n^2) won't land me the job.
I would imagine a hashmap of some sort would come into it, but I just don't have the experience and knowledge to know how to use them.
Here is the code:
function pairwise(arr, arg) {
let usedIndex = [];
let output = 0;
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
if (
arr[i] + arr[j] == arg
&& usedIndex.indexOf(i) == -1
&& usedIndex.indexOf(j) == -1
) {
usedIndex.push(i, j)
output += i + j
}
}
}
return output;
}
pairwise([0, 0, 0, 0, 1, 1], 1) // should return 10
This can be done in one loop with some clever use of an object and knowledge that indices can only be used once.
function pairwise(arr, arg) {
let map = {};
let output = 0;
let length = arr.length;
for (let i = 0; i < length; i++) {
let subArr = map[arr[i]];
if(subArr && subArr[0] !== undefined) {
//there is an index waiting to pair, remove it and add to output
output += subArr.pop() + i;
} else {
//add this index to its pair slot
let left = arg - arr[i];
if(!map[left]) map[left] = [];
map[left].unshift(i);
}
}
return output;
}
console.log(pairwise([0, 0, 0, 0, 1, 1], 1), "should be 10");
console.log(pairwise([1, 4, 2, 3, 0, 5], 7), "should be 11");
console.log(pairwise([], 100), "should be 0");
console.log(pairwise([1, 3, 2, 4], 4), "should be 1");
The keys of the map represent the other value needed to make a pair, and the values of the map are arrays of indices that have the value that would make a pair. The indices are inserted using unshift() so that pop() returns the first one that was inserted - the smallest one.
Iterating from the front guarantees that the smallest pairs are found first, and the map guarantees that any later index will know exactly what the earliest index that could make a pair is.
For a better performance you can save the arr.length into a variable, then for loop won't count every single loop.
function pairwise(arr, arg) {
let usedIndex = [];
let output = 0;
let length = arr.length;
for (let i = 0; i < length; i++) {
for (let j = i + 1; j < length; j++) {
if (
arr[i] + arr[j] == arg
&& usedIndex.indexOf(i) == -1
&& usedIndex.indexOf(j) == -1
) {
usedIndex.push(i, j)
output += i + j
}
}
}
return output;
}
Sort the list.
Have two counters walking from the ends. At each step check to see if the sum is what you want. If so, capture the desired metric.
Step 1 is O(n*logn).
Step 2 is O(n) -- each counter will go about halfway through the list, stopping when they meet.
Write a function that takes in a non-empty array of distinct integers and a target integer.
Your function should find all triplets in the array that sum up to the target sum and return a two-dimensional array of all these triplets.
Each inner array containing a single triplet should have all three of its elements ordered in ascending order
ATTEMPT
function threeNumberSum(arr, target) {
let results = [];
for (let i = 0; i < arr.length; i++) {
let finalT = target - arr[i];
let map = {};
for (let j = i+1; j < arr.length; j++) {
if (map[arr[j]]) {
results.push([arr[j], arr[i], map[arr[j]]]);
} else {
map[finalT-arr[j]] = arr[j];
}
}
}
return results;
}
My code is formatted all funny, but right now im not getting any output. Am I missing a console log somewhere or something?
Your problem is that you read input wrong.
Pay attention to the last part of question: How to Read Input that is Used to Test Your Implementation
You wrote a function that takes the array as first arg and the target integer as the second one. But the input is entered one by one, so your program should read one value at a time from the console input.
I have a string "ABCDEFGHIJKLMN" that I need to shuffle in a specific manner. To do that, I write the characters sequentially in columns bottom -> top and then left -> right (4 chars per column for example) until all characters are done. If the last column is not complete, then the empty spaces need to be on the bottom (this is very important). Like so:
D H L N
C G K M
B F J
A E I
The shuffle is accomplished by producing a new string reading the block of letters as we read text, in rows left -> right:
"DHLNCGKMBFJAEI"
The cases where the columns are not complete (word.size % column_height !=0) complicate things considerably.
I came up with a few solutions, but I'm not sure if there is a simpler (ie, shorter OR easier to read) and more elegant way of coding this problem. My solutions either have an ugly, separate block of code to handle the final incomplete column or seem way too complicated.
My question is, could it be done better?
If you don't want any spoilers and decide to try and figure it out for yourself, stop reading now. If you want to work from what I fiddled so far, then a working piece of code is
var result = "";
var str = "ABCDEFGHIJKLMN";
var nr_rows = 4;
var current_row = 4;
var columns = Math.floor(str.length / nr_rows);
var modulus_table = str.length % nr_rows;
var modulus_position = -1;
for (var i = 0; i < nr_rows; i++) {
for (var j = 0; j < columns; j++) {
result += str[current_row + j * nr_rows - 1];
}
if (modulus_table > 0) {
result += str[str.length + modulus_position];
modulus_table--;
modulus_position--;
}
current_row--;
}
console.log(result);
Moving on to arrays, the next example would loop through each character, placing it correctly in a matrix-like array, but it doesn't work. The array needs to be created another way. For another example of this issue, see How to create empty 2d array in javascript?. This would also need an ugly hack to fix the last characters on the last incomplete column aligning to the bottom instead of the top.
var result = [[],[]];
var str = "ABCDEFGHIJKLMN";
var nr_rows = 4;
var row = nr_rows - 1;
var column = 0;
for (var i = 0; i < str.length; i++) {
result[row][column] = str[i];
row--;
if (row < 0) {
row = nr_rows;
column++;
}
}
console.log(result);
This last method goes full matrix array, but it quickly becomes complicated, since it needs to loop through the array in 3 different directions. First, create a dummy array with the characters in the wrong place, but where the 'undefined' positions correspond to those that should be left empty. That is acomplished by populating the array 'rotated 90º' from the reading orientation.
Without this first step, the empty positions would be stacked at the bottom instead of the top.
A second pass is required to re-write the caracters in the correct places, skipping any holes in the matrix using the 'undefined' value. This check is made for every position and there is no separate block of code to handle an incomplete last line.
A third pass then reads every character in order to form the final shuffled string. All this seems way too complicated and confusing.
// matrix populated top->bottom and left->right
// with the characters in the wrong place
// but the undefined postions in the correct place of the empty positions
var matrix = [];
var str = "ABCDEFGHIJKLMN";
var rows = 4;
var columns = Math.ceil(str.length / rows);
var k = 0;
for (var i = 0; i < rows; i++) {
matrix[i] = [];
for (var j = columns - 1; j >= 0; j--) {
matrix[i][j] = str[k];
k++;
}
}
// populate the matrix with the chars in the correct place and the 'undefined' positions left empty
var k = 0;
for (var i = 0; i < rows; i++) {
for (var j = 0; j < columns; j++) {
if (matrix[i][j] != undefined) {
matrix[i][j] = str[k];
k++;
}
}
}
// read matrix in correct direction and send to string, skipping empty positions
var result = "";
for (var j = columns - 1; j >= 0; j--) {
for (var i = 0; i < rows; i++) {
if (matrix[i][j] != undefined) {
result += matrix[i][j];
}
}
}
console.log(result);
What if you just split/reverse the array into column groups, and convert to rows?
const result = str.match(/.{1,4}/g) // split string into groups of 4
.map(i => i.split('').reverse()) // reverse each group (bottom to top, and solves the last col issue)
.reduce((res, col) => { // reduce the groups into rows
col.forEach((c, i) => res[i] += c) // concat each string char to the right row
return res
}, ['','','','']) // initialise empty strings per row
.join('') // join the rows up
Fiddle here
If you wish to return a string, I don't see why any intermediate result should use an array when it doesn't have to. The following could use one less array, but it's convenient to use split and an array to control the while loop rather than mutate the string.
The idea is to fill the strings from the bottom up until the column is full, then keep adding from the bottom of each column until it runs out of characters to assign. The row to start filling from is based on how many characters are left and how many rows there are.
Rather than building strings, it could build arrays but then generating a string requires multiple joins.
It can also produce results where there are insufficient slots for all the characters, so a result using 9 characters from 10 or more using a 3x3 "matrix" (see last example).
function verticalShuffle(s, rows, cols) {
var result = [''];
s = s.split('');
while (s.length && result[0].length < cols) {
for (var i = (rows < s.length? rows : s.length) -1 ; i>=0; i--) {
if (!result[i]) result[i] = '';
result[i] += s.splice(0,1)[0] || '';
}
}
return result.join('');
}
var s = 'ABCDEFGHIJKLMN';
console.log(verticalShuffle(s, 4, 4)); // DHLNCGKMBFJAEI
console.log(verticalShuffle(s, 6, 3)); // FLNEKMDJCIBHAG
// Only use 9 characters
console.log(verticalShuffle(s, 3, 3)); // CFIBEHADG
This uses plain ed3 functionality that will run in any browser. I don't see the point of restricting it to ECMAScript 2015 or later hosts.
If interpret Question correctly, you can use for loop, String.prototype.slice() to to populate arrays with characters of string. Use Array.prototype.pop() within recursive function to get last element of array until each array .length is 0.
To create array
[
["D","H","L","N"],
["C","G","K","M"],
["B","F","J"],
["A","E","I"]
]
from vertically inverted string you can use for loop, String.prototype.slice() to set array of arrays containing elements having .length 4, or 3 once .length of parent array is 2, having been set with two arrays containing four elements
var str = "ABCDEFGHIJKLMN";
function fnVerticalInvert(str, arr, res) {
if (!arr && !res) {
arr = []; res = "";
}
if (str) {
for (var i = 0; i < str.length; i += 4) {
arr.push([].slice.call(str.slice(i, i + 4)));
}
}
for (var i = 0; i < arr.length; i++) {
if (arr[i].length) {
res += arr[i].pop()
}
}
if (arr.some(function(curr) {return curr.length}))
return fnVerticalInvert(null, arr, res);
for (var i = 0, l = 4, j = 0, n = l - 1, k; i < l; i++, j += l) {
if (i === l / 2) k = j;
arr[i] = [].slice.call(res.slice(!k ? j : k, !k ? j + l : k + n));
if (k) k += n;
}
return {str: res, arr:arr};
};
var res = fnVerticalInvert(str);
console.log(res);
I have a comma separated string, out of which I need to create a new string which contains a random order of the items in the original string, while making sure there are no recurrences.
For example:
Running 1,2,3,1,3 will give 2,3,1 and another time 3,1,2, and so on.
I have a code which picks a random item in the original string, and then iterates over the new string to see if it does not exist already. If it does not exist - the item is inserted.
However, I have a feeling this can be improved (in C# I would have used a hashtable, instead of iterating every time on the new array). One improvement can be removing the item we inserted from the original array, in order to prevent cases where the random number will give us the same result, for example.
I'd be happy if you could suggest improvements to the code below.
originalArray = originalList.split(',');
for (var j = 0; j < originalArray.length; j++) {
var iPlaceInOriginalArray = Math.round(Math.random() * (originalArray.length - 1));
var bAlreadyExists = false;
for (var i = 0; i < newArray.length; i++) {
if (newArray[i].toString() == originalArray[iPlaceInOriginalArray].toString()) {
bAlreadyExists = true;
break;
}
}
if (!bAlreadyExists)
newArray.push(originalArray[iPlaceInOriginalArray]);
}
Thanks!
You can still use a 'hash' in javascript to remove duplicates. Only in JS they're called objects:
function removeDuplicates(arr) {
var hash = {};
for (var i=0,l=arr.length;i<l;i++) {
hash[arr[i]] = 1;
}
// now extract hash keys... ahem...
// I mean object members:
arr = [];
for (var n in hash) {
arr.push(n);
}
return arr;
}
Oh, and the select random from an array thing. If it's ok to destroy the original array (which in your case it is) then use splice:
function randInt (n) {return Math.floor(Math.random()*n)}
function shuffle (arr) {
var out = [];
while (arr.length) {
out.push(
arr.splice(
randInt(arr.length),1 ));
}
return out;
}
// So:
newArray = shuffle(
removeDuplicates(
string.split(',') ));
// If you sort the first array, it is quicker to skip duplicates, and you can splice each unique item into its random position as you build the new array.
var s= 'Function,String,Object,String,Array,Date,Error,String,String,'+
'Math,Number,RegExp,Group,Collection,Timelog,Color,String';
var A1= s.split(',').sort(), A2= [], tem;
while(A1.length){
tem= A1.shift();
while(A1[0]== tem) tem= A1.shift();
if(tem) A2.splice(Math.floor(Math.random()*A2.length), 0, tem);
}
alert(A2.join(', '))
With your solution, you are not guaranteed not to pick same number several times, thus leaving some others of them never being picked. If the number of elements is not big (up to 100), deleting items from the source array will give the best result.
Edit
originalArray = originalList.split(',');
for (var j = 0; j < originalArray.length; j++) {
var iPlaceInOriginalArray = Math.round(Math.random() * (originalArray.length - 1 - j));
var bAlreadyExists = false;
for (var i = 0; i < newArray.length; i++) {
if (newArray[i].toString() == originalArray[iPlaceInOriginalArray].toString()) {
bAlreadyExists = true;
break;
}
}
var tmp = originalArray[originalArray.length - 1 - j];
originalArray[originalArray.Length - 1 - j] = originalArray[iPlaceInOriginalArray];
originalArray[iPlaceInOriginalArray] = tmp;
if (!bAlreadyExists)
newArray.push(originalArray[iPlaceInOriginalArray]);
}